Legendre's equation and Legendre polynomials (intro): Worked examples

4.6.20 · D3 · Maths › Ordinary Differential Equations › Legendre's equation and Legendre polynomials (intro)

Shuru karne se pehle, notation ka ek reminder, taaki kuch bina explanation ke use na ho:

The scenario matrix

Cell	Case class	Kya special hai	Example
A	Recurrence se naya $P_{n}$ banana	engine test karta hai, even parity	Ex 1 ( $P_{4}$ )
B	Kisi specific $n$ ke liye Rodrigues' formula	alternate route, odd parity	Ex 2 ( $P_{5}$ )
C	Degenerate input $n = 0$	"flat pattern", trivial recurrence	Ex 3
D	Endpoint / sign values $x = \pm 1$ , parity	boundary + negative argument	Ex 4
E	$m \neq = n$ ke saath Orthogonality (different indices)	integral zero hai	Ex 5
F	$m = n$ ke saath Orthogonality (normalization integral)	integral nonzero $= \frac{2}{2 n + 1}$ hai	Ex 6
G	Di gayi polynomial $f (x)$ ko $P_{n}$ basis mein expand karna	real "Fourier-in-polynomials" use	Ex 7
H	Word problem: sphere par potential ( $x = cos θ$ )	physics, poles par limiting behaviour	Ex 8
I	Exam twist: non-integer $n$ , polynomial kyun nahi	degenerate/limiting termination fail hoti hai	Ex 9

Recurrence engine jis par hum baar baar lean karte hain:

Cell A — Recurrence se $P_{4}$ banana

Step 1 — parity chunna. Kyunki $n = 4$ even hai, even series use karo: $a_{0}$ se shuru karo, $a_{1} = 0$ set karo. Yeh step kyun? Recurrence coefficients ko do apart link karta hai; $n = 4$ even branch ko degree 4 par terminate karta hai, isliye odd branch sirf garbage add karega (aur ek genuine $P_{n}$ ke liye vanish hona chahiye).

Step 2 — ladder chadhna. $n = 4$ ke saath, $n (n + 1) = 20$ . $m = 0, 2, 4$ ko engine mein feed karo:

$a_{2} = \frac{0 \cdot 1 - 20}{2 \cdot 1} a_{0} = \frac{- 20}{2} a_{0} = - 10 a_{0} .$
$a_{4} = \frac{2 \cdot 3 - 20}{4 \cdot 3} a_{2} = \frac{- 14}{12} a_{2} = - \frac{7}{6} (- 10 a_{0}) = \frac{35}{3} a_{0} .$
$a_{6} = \frac{4 \cdot 5 - 20}{6 \cdot 5} a_{4} = \frac{0}{30} a_{4} = 0.$ ✔ terminate ho gaya.

Yeh step kyun? Har rung pichle wale ko use karta hai; $m = n = 4$ par numerator ka $0$ hit karna exactly "magic termination" hai — yeh degree $6$ aur upar ki sab cheez ko kill kar deta hai.

Step 3 — assemble karo. $y = a_{0} (1 - 10 x^{2} + \frac{35}{3} x^{4}) .$

Step 4 — $P_{4} (1) = 1$ se normalize karo. $a_{0} (1 - 10 + \frac{35}{3}) = a_{0} \cdot \frac{3 - 30 + 35}{3} = a_{0} \cdot \frac{8}{3} = 1 \Rightarrow a_{0} = \frac{3}{8} .$ Toh $P_{4} (x) = \frac{3}{8} (1 - 10 x^{2} + \frac{35}{3} x^{4}) = \frac{1}{8} (35 x^{4} - 30 x^{2} + 3) .$

Verify: $P_{4} (1) = \frac{1}{8} (35 - 30 + 3) = \frac{1}{8} \cdot 8 = 1$ ✔. Parent ke forecast answer se match karta hai. Top coefficient positive — apna guess check karo.

Cell B — $P_{5}$ ke liye Rodrigues' formula (odd parity)

Step 1 — prefactor likhna. $2^{5} \cdot 5! = 32 \cdot 120 = 3840$ . Yeh step kyun? Yeh constant hi hai jo $P_{n} (1) = 1$ sahi nikalti hai; isko galat karna poore answer ko scale kar deta hai.

Step 2 — binomial theorem se $(x^{2} - 1)^{5}$ expand karna. Binomial theorem kehta hai $(a + b)^{k} = \sum_{j = 0}^{k} (j k) a^{k - j} b^{j}$ ; yahan $a = x^{2}$ , $b = - 1$ , $k = 5$ , aur coefficients $(j 5)$ Pascal's row $1, 5, 10, 10, 5, 1$ se read off hote hain $(- 1)^{j}$ se alternating signs ke saath: $(x^{2} - 1)^{5} = x^{10} - 5 x^{8} + 10 x^{6} - 10 x^{4} + 5 x^{2} - 1.$ Yeh step kyun? Ek clean polynomial ko 5 baar differentiate karna product differentiate karne se kahin asaan hai; binomial theorem ek line mein woh clean expansion deta hai.

Step 3 — paanch baar differentiate karna. Term-by-term, $\frac{d ^{5}}{d x ^{5}} x^{k} = k (k - 1) (k - 2) (k - 3) (k - 4) x^{k - 5}$ (aur $0$ agar $k < 5$ ):

$x^{10} \to 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 x^{5} = 30240 x^{5}$
$- 5 x^{8} \to - 5 (8 \cdot 7 \cdot 6 \cdot 5 \cdot 4) x^{3} = - 5 \cdot 6720 x^{3} = - 33600 x^{3}$
$10 x^{6} \to 10 (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2) x = 10 \cdot 720 x = 7200 x$
neeche ke terms vanish ho jaate hain.

Yeh step kyun? Sirf degree $\geq 5$ ke terms paanch derivatives ke baad survive karte hain — yahi wajah hai ki result ki degree exactly $n = 5$ hoti hai.

Step 4 — $3840$ se divide karna. $P_{5} (x) = \frac{30240 x ^{5} - 33600 x ^{3} + 7200 x}{3840} = \frac{1}{8} (63 x^{5} - 70 x^{3} + 15 x) .$

Verify: $P_{5} (1) = \frac{1}{8} (63 - 70 + 15) = \frac{1}{8} \cdot 8 = 1$ ✔. Aur $P_{5} (0) = 0$ ✔ (odd), forecast se match karta hai.

Cell C — Degenerate case $n = 0$

Step 1 — $n = 0$ ke saath equation likhna. $(1 - x^{2}) y^{''} - 2 x y^{'} = 0.$ Yeh step kyun? $n = 0$ plug karne se undifferentiated term hat jata hai, aur $u = y^{'}$ mein ek first-order equation bachi rehti hai.

Step 2 — $u = y^{'}$ substitute karna. $(1 - x^{2}) u^{'} - 2 xu = 0 \Rightarrow \frac{u ^{'}}{u} = \frac{2 x}{1 - x ^{2}} .$ Yeh step kyun? Ab yeh separable hai; $\frac{2 x}{1 - x ^{2}}$ exactly $- \frac{d}{d x} ln (1 - x^{2})$ hai, jise hum integrate kar sakte hain.

Step 3 — integrate karna. $ln u = - ln (1 - x^{2}) + C \Rightarrow u = \frac{K}{1 - x ^{2}} .$ Phir $y = K \int \frac{d x}{1 - x ^{2}} = \frac{K}{2} ln \frac{1 + x}{1 - x} + D .$

Step 4 — poles par boundedness apply karna. Jab $x \to \pm 1$ , $ln \frac{1 + x}{1 - x} \to \pm \infty$ . Bounded rehne ke liye hume $K = 0$ set karna hoga, aur $y = D =$ const bachega. $P_{0} (1) = 1$ normalize karo: $D = 1$ . $P_{0} (x) = 1.$

Verify: Discard ki gayi piece, constant factor tak, $artanh (x) = \frac{1}{2} ln \frac{1 + x}{1 - x}$ hai, jiska derivative $\frac{1}{1 - x ^{2}}$ hai — $u$ se match karta hai. Yeh $Q_{0} (x)$ hai, log-divergent second solution jiske baare mein parent ne warn kiya tha. Forecast ka answer: haan, ek doosra solution exist karta hai — hum bas use unbounded hone ki wajah se reject kar dete hain.

Cell D — Endpoint values aur negative argument

Step 1 — $x = - 1$ par parity use karna. Rule $P_{n} (- 1) = (- 1)^{n}$ hai.

$P_{3} (- 1) = (- 1)^{3} = - 1.$
$P_{4} (- 1) = (- 1)^{4} = + 1.$

Yeh step kyun? $P_{n} (- 1) = P_{n} (- x) ∣_{x = 1} = (- 1)^{n} P_{n} (1) = (- 1)^{n}$ . Parity free mein kaam karta hai — koi polynomial evaluation ki zaroorat nahi.

Step 2 — poore function ko reflect karna. $P_{3} (- x) = (- 1)^{3} P_{3} (x) = - P_{3} (x)$ . Explicitly $- \frac{1}{2} (5 x^{3} - 3 x) = \frac{1}{2} (3 x - 5 x^{3})$ — aur indeed $P_{3} (- x) = \frac{1}{2} (5 (- x)^{3} - 3 (- x)) = \frac{1}{2} (- 5 x^{3} + 3 x)$ ✔.

Step 3 — ek plain interior value. $P_{2} (x) = \frac{1}{2} (3 x^{2} - 1)$ , toh $P_{2} (- \frac{1}{2}) = \frac{1}{2} (3 \cdot \frac{1}{4} - 1) = \frac{1}{2} (\frac{3}{4} - 1) = \frac{1}{2} \cdot (- \frac{1}{4}) = - \frac{1}{8} .$

Verify: Evenness se $P_{2} (- \frac{1}{2}) = P_{2} (\frac{1}{2}) = \frac{1}{2} (\frac{3}{4} - 1) = - \frac{1}{8}$ ✔. Forecast: $P_{4} (- 1) = + 1$ , $P_{3} (- 1) = - 1$ .

Cell E — Different indices ke saath Orthogonality ( $m \neq = n$ )

Step 1 — orthogonality invoke karna. $\int_{- 1}^{1} P_{m} P_{n} d x = \frac{2}{2 n + 1} δ_{mn}$ . Yahan $m = 2 \neq = n = 3$ , toh $δ_{mn} = 0$ . Yeh step kyun? Ek Sturm–Liouville problem ke different eigenvalues ⇒ orthogonal eigenfunctions (dekho Sturm-Liouville Theory). Koi computation required nahi.

Step 2 — parity se sanity check. $P_{2}$ even hai, $P_{3}$ odd hai, toh product odd hai; symmetric interval $[- 1, 1]$ par ek odd function ka integral $0$ hota hai. Yeh step kyun? Yeh answer ko ek independent route se confirm karta hai — parity, orthogonality nahi.

$\int_{- 1}^{1} P_{2} P_{3} d x = 0.$

Verify: Direct: $\int_{- 1}^{1} \frac{1}{2} (3 x^{2} - 1) \cdot \frac{1}{2} (5 x^{3} - 3 x) d x$ ek odd polynomial integrate karta hai → $0$ ✔.

Cell F — Equal indices ke saath Orthogonality (normalization integral)

Step 1 — $P_{2}$ ko square karna. $[\frac{1}{2} (3 x^{2} - 1)]^{2} = \frac{1}{4} (9 x^{4} - 6 x^{2} + 1)$ . Yeh step kyun? Theorem check karne ke liye hume explicit integrand chahiye, sirf quote nahi karna.

Step 2 — har power ko $[- 1, 1]$ par integrate karna. $\int_{- 1}^{1} x^{2 k} d x = \frac{2}{2 k + 1}$ use karke: $\int_{- 1}^{1} x^{4} = \frac{2}{5}, \int_{- 1}^{1} x^{2} = \frac{2}{3}, \int_{- 1}^{1} 1 = 2.$

Step 3 — combine karna. $\frac{1}{4} (9 \cdot \frac{2}{5} - 6 \cdot \frac{2}{3} + 2) = \frac{1}{4} (\frac{18}{5} - 4 + 2) = \frac{1}{4} (\frac{18}{5} - 2) = \frac{1}{4} \cdot \frac{8}{5} = \frac{2}{5} .$

Verify: Rule deta hai $\frac{2}{2 ( 2 ) + 1} = \frac{2}{5}$ ✔. Match karta hai. Aur $\frac{2}{5} > 0$ bhi hai, jaise kisi bhi square ke integral ka hona chahiye.

Cell G — Di gayi polynomial ko $P_{n}$ basis mein expand karna

Step 1 — decide karna ki kaun se $P_{n}$ appear kar sakte hain. Degree $\leq 2$ aur even parity ⇒ sirf $P_{0} = 1$ aur $P_{2} = \frac{1}{2} (3 x^{2} - 1)$ . Yeh step kyun? $P_{n}$ ek basis banate hain; ek degree-2 even polynomial poori tarah ${P_{0}, P_{2}}$ ke span mein rehti hai, toh sum finite hai.

Step 2 — algebra se match karna (yahan sabse fast). $x^{2} = c_{0} P_{0} + c_{2} P_{2} = c_{0} + c_{2} \frac{1}{2} (3 x^{2} - 1)$ likhna. $x^{2}$ ka coefficient: $\frac{3}{2} c_{2} = 1 \Rightarrow c_{2} = \frac{2}{3}$ . Constant: $c_{0} - \frac{1}{2} c_{2} = 0 \Rightarrow c_{0} = \frac{1}{2} c_{2} = \frac{1}{3}$ .

Step 3 — projection formula se cross-check karna. $c_{n} = \frac{2 n + 1}{2} \int_{- 1}^{1} x^{2} P_{n} d x$ . $c_{0} = \frac{1}{2} \int_{- 1}^{1} x^{2} d x = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}, c_{2} = \frac{5}{2} \int_{- 1}^{1} x^{2} \cdot \frac{1}{2} (3 x^{2} - 1) d x .$ Integral $= \frac{1}{2} (3 \cdot \frac{2}{5} - \frac{2}{3}) = \frac{1}{2} (\frac{6}{5} - \frac{2}{3}) = \frac{1}{2} \cdot \frac{8}{15} = \frac{4}{15}$ hai, toh $c_{2} = \frac{5}{2} \cdot \frac{4}{15} = \frac{2}{3}$ ✔. Yeh step kyun? Orthogonality "coefficients dhundho" ko har ek ke liye ek single integral mein badal deta hai — Fourier Series idea polynomial clothing mein.

$x^{2} = \frac{1}{3} P_{0} (x) + \frac{2}{3} P_{2} (x) .$

Verify: $\frac{1}{3} \cdot 1 + \frac{2}{3} \cdot \frac{1}{2} (3 x^{2} - 1) = \frac{1}{3} + x^{2} - \frac{1}{3} = x^{2}$ ✔.

Cell H — Word problem: ek hemisphere par potential (physics, $x = cos θ$ )

Figure — Legendre's equation and Legendre polynomials (intro)

Figure kaise padhein. Left panel: ek sphere edge-on dekha gaya; har coloured spoke north pole se angle $θ$ par draw ki gayi hai, aur uska colour wahan surface potential $V$ encode karta hai — top ke paas warm (high $V$ ) neeche ki taraf cool hota hua. Do labelled arrows un poles ko mark karte hain jahan hum values read karenge. Right panel: wahi $V$ $θ$ ke against plot kiya gaya; $θ = 0$ par coral dot $θ = π$ par lavender dot se kaafi upar hai — yeh ek visual proof hai ki top aur bottom same nahi hain, jo exactly odd $P_{1}$ (dipole) term ki fingerprint hai. Agar $V$ mein sirf even modes hote, toh curve $θ = π /2$ ke baare mein symmetric hoti; jo tilt tum dekhte ho woh dipole hai.

Step 1 — $x = cos θ$ substitute karna. $V = 1 + 2 x + \frac{3}{2} x^{2}$ . Yeh step kyun? Laplace's Equation in Spherical Coordinates mein, boundary data naturally $cos θ$ ki powers mein aata hai; $x$ mein convert karna hume directly Legendre machinery use karne deta hai.

Step 2 — Ex 7 ka $x^{2}$ expansion use karna. $x^{2} = \frac{1}{3} P_{0} + \frac{2}{3} P_{2}$ , plus $x = P_{1}$ , $1 = P_{0}$ : $V = P_{0} + 2 P_{1} + \frac{3}{2} (\frac{1}{3} P_{0} + \frac{2}{3} P_{2}) = (1 + \frac{1}{2}) P_{0} + 2 P_{1} + P_{2} .$ $V = \frac{3}{2} P_{0} + 2 P_{1} + P_{2} .$ Yeh step kyun? Har $P_{n}$ ek "multipole mode" hai; coefficients physical monopole/dipole/quadrupole strengths hain.

Step 3 — $P_{n} (\pm 1)$ use karke poles par evaluate karna. North pole $x = 1$ : $V = \frac{3}{2} (1) + 2 (1) + 1 = \frac{9}{2} = 4.5$ . South pole $x = - 1$ : $V = \frac{3}{2} (1) + 2 (- 1) + 1 (1) = \frac{3}{2} - 2 + 1 = \frac{1}{2} = 0.5$ .

Verify: $V = 1 + 2 x + \frac{3}{2} x^{2}$ se directly: $x = 1$ par, $1 + 2 + 1.5 = 4.5$ ✔; $x = - 1$ par, $1 - 2 + 1.5 = 0.5$ ✔. Do poles differ karte hain — $P_{1}$ (dipole, top-vs-bottom) term symmetry torti hai, exactly wahi jo figure ki asymmetric shading dikhati hai. Dono values finite hain — $P_{n}$ poles par bounded hote hain, physical selection rule kaam mein.

Cell I — Exam twist: non-integer $n$ , polynomial kyun nahi

Step 1 — $n (n + 1)$ likhna. $n (n + 1) = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4}$ . Yeh step kyun? Termination ke liye koi integer $m$ chahiye jisme $m (m + 1) = \frac{3}{4}$ ho. Lekin $m (m + 1)$ integer $m$ ke liye hamesha integer hota hai ( $0, 2, 6, 12, \dots$ ), kabhi $\frac{3}{4}$ nahi.

Step 2 — pehle kuch ratios compute karna (even branch $a_{0} \neq = 0$ ).

$a_{2} = \frac{0 - \frac{3}{4}}{2} a_{0} = - \frac{3}{8} a_{0} .$
$a_{4} = \frac{2 \cdot 3 - \frac{3}{4}}{12} a_{2} = \frac{21/4}{12} a_{2} = \frac{21}{48} a_{2} = \frac{7}{16} \cdot (- \frac{3}{8}) a_{0} = - \frac{21}{128} a_{0} .$
$a_{6} = \frac{4 \cdot 5 - \frac{3}{4}}{30} a_{4} \neq = 0$ — aur yeh kabhi zero nahi hota.

Yeh step kyun? $a_{6} \neq = 0$ explicitly dekhna prove karta hai ki series genuinely infinite hai; koi clean stopping index nahi hai.

Step 3 — endpoints par convergence argument (ratio test). Bade $m$ ke liye successive even coefficients ka ratio dekho: $\frac{a _{m + 2}}{a _{m}} = \frac{m ( m + 1 ) - \frac{3}{4}}{( m + 2 ) ( m + 1 )} m \to \infty \frac{m ^{2}}{m ^{2}} = 1.$ Zyada precisely ratio $\frac{m}{m + 2} = 1 - \frac{2}{m} + \dots$ ki tarah behave karta hai, toh coefficients sirf $1/ m$ ki tarah shrink hote hain — bahut zyada slow. $x = 1$ par divergent series $\sum \frac{1}{k}$ se compare karo: $x = 1$ par terms $a_{m} x^{m}$ itni fast zero ki taraf nahi jaate ki summable ho saken, toh series $x \to \pm 1$ par diverge karti hai. Yeh step kyun? Yahi wajah hai ki physics $n$ ko non-negative integer hone par force karti hai — tab hi numerator vanish hota hai, series rukti hai, aur solution poles par finite rehta hai.

Verify: Numerically, $a_{4} / a_{0} = - \frac{21}{128} \approx - 0.1641$ hai, aur ratio $\frac{a _{m + 2}}{a _{m}} = \frac{m ( m + 1 ) - 3/4}{( m + 2 ) ( m + 1 )} \to 1$ as $m \to \infty$ — boundary-divergent series ki tell-tale. Non-integer $n$ ⇒ koi polynomial nahi ✔.

Recall Self-test: example ko matrix cell se match karo

Kaun sa cell hai "answer parity se zero hai"? ::: Cell E ( $m \neq = n$ orthogonality). Kaun sa cell second solution $Q_{0}$ explicitly dikhata hai? ::: Cell C ( $n = 0$ degenerate case). Kaun sa cell prove karta hai ki non-integer $n$ koi polynomial nahi deta? ::: Cell I. Kaun se coefficients $x^{2}$ express karte hain? ::: $x^{2} = \frac{1}{3} P_{0} + \frac{2}{3} P_{2}$ (Cell G).

Connections

Power Series / Frobenius Method — recurrence engine jo Cells A, C, I mein use hota hai.
Sturm-Liouville Theory — Cells E, F ki orthogonality kyun hold karti hai.
Laplace's Equation in Spherical Coordinates — Cell H ke potential problem ka origin.
Fourier Series — Cell G ka "basis mein expand karo" idea.
Generating Functions — same $P_{n}$ tak ek alternate route.
Associated Legendre Functions & Spherical Harmonics — is intro ke baad ka agla step.